CHAPTER 6
Application and Experimental Questions
E1.
A variegated trait in plants is analyzed using reciprocal crosses. The following
results are obtained:
Variegated female × Normal male
Normal female × Variegated male
↓
↓
1,024 variegated + 52 normal
1,113 normal + 61 variegated
Explain this pattern of inheritance.
Answer: Based on these results, the pattern of inheritance may be cytoplasmic
inheritance involving plastids. In most cases, plastids follow a maternal inheritance
pattern, but a low percentage of the time, paternal leakage does occur.
E2.
Ruth Sager and her colleagues discovered that the mode of inheritance of
streptomycin resistance in Chlamydomonas could be altered if the mt+ cells were
exposed to UV irradiation prior to mating. This exposure dramatically increased the
frequency of biparental inheritance. What would be the expected outcome of a cross
between an mt+ smr and an mt– sms strain in the absence of UV irradiation? How would
the result differ if the mt+ strain was exposed to UV light?
Answer: In the absence of UV light, we would expect all smr offspring. With UV light,
we would expect a greater percentage of sms offspring.
E3.
Two male mice, which we will call male A and male B, are both phenotypically
normal. Male A was from a litter that contained half phenotypically normal mice and half
dwarf mice. The mother of male A was known to be homozygous for the normal Igf2
allele. Male B was from a litter of eight mice that were all phenotypically normal. The
parents of male B were a phenotypically normal male and a dwarf female. Male A and
male B were put into a cage with two female mice that we will call female A and female
B. Female A is dwarf, and female B is phenotypically normal. The parents of these two
females were unknown, although it was known that they were from the same litter. The
mice were allowed to mate with each other and the following data were obtained:
Female A gave birth to three dwarf babies and four normal babies.
Female B gave birth to four normal babies and two dwarf babies.
Which male(s) mated with female A and female B? Explain.
Answer: Let’s first consider the genotypes of male A and male B. Male A must have two
normal copies of the Igf2 gene. We know this because male A’s mother was Igf2 Igf2; the
father of male A must have been a heterozygote Igf2 Igf2‒, because half of the litter that
contained male A also contained dwarf offspring. But because male A was not dwarf, it
must have inherited the normal allele from its father. Therefore, male A must be Igf2
Igf2. We cannot be completely sure of the genotype of male B. It must have inherited the
normal Igf2 allele from its father because male B is phenotypically normal. We do not
know the genotype of male B’s mother, but she could be either Igf2‒ Igf2‒ or Igf2 Igf2‒.
In either case, the mother of male B could pass the Igf2‒ allele to an offspring, but we do
not know for sure if she did. So, male B could be either Igf2 Igf2‒ or Igf2 Igf2.
For the Igf2 gene, we know that the maternal allele is inactivated. Therefore, the
genotypes and phenotypes of females A and B are irrelevant. The phenotype of the
offspring is determined only by the allele that is inherited from the father. Because we
know that male A has to be Igf2 Igf2, we know that it can produce only normal offspring.
Because both females A and B both produced dwarf offspring, male A cannot be the
father. In contrast, male B could be either Igf2 Igf2 or Igf2 Igf2‒. Because both females
gave birth to dwarf babies (and because male A and male B were the only two male mice
in the cage), we conclude that male B must be Igf2 Igf2‒ and is the father of both litters.
E4.
Figure 6.14 describes an example of a maternal effect gene. Explain how
Sturtevant deduced a maternal effect gene based on the F2 and F3 generations.
Answer: The results of each succeeding generation depended on the genotypes of the
mothers of the preceding generation. For example, if a mother was dd, the F1 offspring
were all sinistral. The genotypes of the F2 mothers were 1 DD : 2 Dd : 1 dd. The DD and
Dd mothers produced dextral offspring of the F3 generation and the dd mothers produced
sinistral offspring. As expected, the ratio of dextral to sinistral was 3:1, which is derived
from the genotypes of the F2 mothers, which were 1 DD : 2 Dd : 1 dd.
E5.
Chapter 20 describes three blotting methods (i.e., Southern blotting, Northern
blotting, and Western blotting) that are used to detect specific genes and gene products.
Southern blotting detects DNA, Northern blotting detects RNA, and Western blotting
detects proteins. Let’s suppose that a female fruit fly is heterozygous for a maternal effect
gene, which we will call gene B. The female is Bb. The normal allele, B, encodes a
functional mRNA that is 550 nucleotides long. A recessive allele, b, encodes a shorter
mRNA that is 375 nucleotides long. (Allele b is due to a deletion within this gene.) How
could you use one or more of these techniques to show that nurse cells transfer gene
products from gene B to developing eggs? You may assume that you can dissect the
ovaries of fruit flies and isolate eggs separately from nurse cells. In your answer, describe
your expected results.
Answer: A haploid egg should express either the 550- or 375-length mRNA, but not both
(because it has only one copy of the gene). The nurse cells, however, can express both
mRNAs if the female is heterozygous. Therefore, if we begin with heterozygous females,
we could dissect and separate the nurse cells from the egg. We could then isolate mRNA
from the nurse cells and (in a separate tube) isolate mRNA from eggs. The mRNA would
then be run on a gel and subjected to Northern blotting, using a probe that is
complementary to both the 550- and 375-length mRNA. According to our knowledge of
maternal effect genes, we would expect the egg to contain both the 550 and 375 mRNAs,
because it receives them from the nurse cells. Both forms of the mRNA would also be
found in the nurse cells.
E6.
As a hypothetical example, a trait in mice results in mice with very long tails. You
initially have a true-breeding strain with normal tails and a true-breeding strain with long
tails. You then make the following types of crosses:
Cross 1: When true-breeding females with normal tails are crossed to true-breeding males
with long tails, all F1 offspring have long tails.
Cross 2: When true-breeding females with long tails are crossed to true-breeding males
with normal tails, all F1 offspring have normal tails.
Cross 3: When F1 females from cross 1 are crossed to true-breeding males with normal
tails, all offspring have normal tails.
Cross 4: When F1 males from cross 1 are crossed to true-breeding females with long
tails, half of the offspring have normal tails and half have long tails.
Explain the pattern of inheritance of this trait.
Answer: The pattern of inheritance is consistent with imprinting. In every cross, the allele
that is inherited from the father is expressed in the offspring, whereas the allele inherited
from the mother is not.
E7.
You have a female snail that coils to the right, but you do not know its genotype.
You may assume that right coiling (D) is dominant to left coiling (d). You also have male
snails at your disposal of known genotype. How would you determine the genotype of
this female snail? In your answer, describe your expected results depending on whether
the female is DD, Dd, or dd.
Answer: Mate the female to a dd male. If all of the offspring coil to the left, you know
the female must be dd. If they all coil to the right, she could be either DD or Dd. If the F1
offspring coil to the right, you could let them mate with each other to produce an F2
generation. If the original mother was Dd, then half of the F1 female offspring would be
Dd and half would be dd. Therefore, half of the F2 snails would coil to the right and half
to the left. In contrast, if the original mother was DD, all of the F1 female offspring
would be Dd. In this case, all of the F2 snails would coil to the right.